If `x = 2015 , y = 2014 and z = 2013,` then the value of `x ^(2) + y^(2) + z ^(2) - xy - yz - zx` is :

Options

A. A)3
B. B)4
C. C)6
D. D)2

Answer

To solve the expression \( x^2 + y^2 + z^2 - xy - yz - zx \) given \( x = 2015 \), \( y = 2014 \), and \( z = 2013 \), we can follow these steps:

### Step 1: Substitute the values of x, y, and z
We start by substituting the values into the expression:
\[ x^2 + y^2 + z^2 - xy - yz - zx = 2015^2 + 2014^2 + 2013^2 - (2015 \cdot 2014) - (2014 \cdot 2013) - (2013 \cdot 2015) \]

### Step 2: Calculate \( x^2, y^2, z^2 \)
Now we calculate each square:
\[ 2015^2 = 4060225 \]
\[ 2014^2 = 4056196 \]
\[ 2013^2 = 4052169 \]

### Step 3: Calculate the product terms
Next, we calculate the products:
\[ 2015 \cdot 2014 = 4058210 \]
\[ 2014 \cdot 2013 = 4054182 \]
\[ 2013 \cdot 2015 = 4052169 \]

### Step 4: Substitute back into the expression
Now we substitute these values back into the expression:
\[ 4060225 + 4056196 + 4052169 - 4058210 - 4054182 - 4052169 \]

### Step 5: Simplify the expression
Now we can simplify the expression step by step:
1. Calculate the sum of squares:
\[ 4060225 + 4056196 + 4052169 = 12118590 \]
2. Calculate the sum of products:
\[ 4058210 + 4054182 + 4052169 = 12104561 \]
3. Now subtract the sum of products from the sum of squares:
\[ 12118590 - 12104561 = 14029 \]

### Step 6: Final calculation
Now, we can see that we can also use a shortcut since the differences between the values of \( x, y, z \) are consistent. The differences are:
\[ x - y = 1, \quad y - z = 1, \quad z - x = -2 \]
Using the formula:
\[ x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2} \left( (x-y)^2 + (y-z)^2 + (z-x)^2 \right) \]
Calculating the differences:
\[ = \frac{1}{2} \left( 1^2 + 1^2 + (-2)^2 \right) = \frac{1}{2} \left( 1 + 1 + 4 \right) = \frac{1}{2} \cdot 6 = 3 \]

Thus, the final value of the expression is:
\[ \boxed{3} \]